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\fluidmechchaptertitle
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\label{chap_three}
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\section{Motivation}

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	\youtubethumb{1LXlFVtPoCY}{pre-lecture briefing for this chapter, part 1/2}{\oc (\ccby)}
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	Our objective for this chapter is to answer the question “what is the \emph{net} effect of a given fluid flow through a given volume?”.
	
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	Here, we develop a mass, momentum and energy accounting methodology to analyze the flow of continuous medium. This method is not powerful enough to allow us to describe extensively the nature of fluid flow around bodies; nevertheless, it is extremely useful to quantify forces, moments, and energy transfers associated with fluid flow.
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\section{The Reynolds transport theorem}

	\subsection{Control volume}
	
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		Let us begin by describing the flow which interests us a a generic velocity field $\vec V = (u, v, w)$ which is a function of space and time ($\vec V = f(x, y, z, t)$).
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		Within this flow, we are interested in an arbitrary volume named \vocab{control volume} (CV) which is free to move and change shape (\cref{fig_cv}). We are going to measure the properties of the fluid at the borders of this volume, which we call the \vocab{control surface} (CS), in order to compute the net effect of the flow through the volume.
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			\begin{figure}
				\begin{center}
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					\includegraphics[width=0.6\textwidth]{concept_control_volume_system.png}
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				\end{center}
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				\supercaption{A control volume within a flow. The \vocab{system} is the amount of mass included within the control volume at a given time. At a later time, it may have left the control volume, and its shape and properties may have changed. The control volume may also change shape with time, although this is not represented here.}{\wcfile{System control volume integral analysis.svg}{Figure} \cczero \oc}
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				\label{fig_cv}
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			\end{figure}
		
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		At a given time, the control volume contains a certain amount of mass which we call the \vocab{system} (sys). Thus the system is a fixed amount of mass transiting through the control volume at the time of our study, and its properties (volume, pressure, velocity etc.) may change in the process.
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		All along the chapter, we are focusing on the question: based on measured fluid properties at some point in space and time (the properties at the control surface), how can we quantify what is happening to the system (the mass inside the control volume)?
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	\subsection{Rate of change of an additive property}
		
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		In order to proceed with our calculations, we need a robust accounting methodology. We start with a “dummy” fluid property $B$, which we will later replace with physical variables of interest.\\
		Let us thus consider an arbitrary additive property $B$ of the fluid. By the term \vocab{additive} property, we mean that the total \emph{amount of property} is divided if the fluid is divided. For instance, this is true of mass, volume, energy, entropy, but not pressure or temperature.\\
		The \vocab{specific} (i.e. per unit mass) value of $B$ is designated $b \equiv B/m$.
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		We now wish to compute the variation of a system’s property $B$ based on measurements made at the borders of the control volume. We will achieve this with an equation containing three terms:
		\begin{itemize}
			\item The time variation of the quantity $B$ within the system is measured with the term $\timederivative{B_\sys}$. This may represent, for example, the rate of change of the fluid’s internal energy as it travels through a jet engine.
			\item Within the control volume, the enclosed quantity $B_\cv$ can vary by accumulation (for example, mass may be increasing in an air tank fed with compressed air): we measure this with the term $\timederivative{B_\cv}$.
			\item Finally, a mass flux may be flowing through the boundaries of the control volume, carrying with it some amount of $B$ every second: we name that net flow out of the system $\dot B_\net \equiv \dot B_\out - \dot B_\inn$.
		\end{itemize}
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		We can now link these three terms with the simple equation:
			\begin{IEEEeqnarray}{CCCCC}
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				\timederivative{B_\sys} & = & \timederivative{B_\cv}	& + & \dot B_\net \label{eq_rtt_basic}\\\nonumber\\
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				\begin{array}{c} {\scriptstyle \text{the rate of change}} \\ {\scriptstyle \text{of $B$ for the system}} \end{array}		& = & \begin{array}{c} {\scriptstyle \text{the rate of change}} \\ {\scriptstyle \text{of $B$ within the}} \\ {\scriptstyle \text{control volume}} \end{array}									& + & \begin{array}{c} {\scriptstyle \text{the net flow of $B$}} \\ {\scriptstyle \text{through the boundaries}} \\ {\scriptstyle \text{of the control volume}} \end{array} \nonumber
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			\end{IEEEeqnarray}
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		Since $B$ may not be uniformly distributed within the control volume, we like to express the term $\timederivative{B_\cv}$ as the integral of the volume density $\frac{B}{\vol}$ with respect to volume:
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			\begin{IEEEeqnarray}{rCl}
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				\timederivative{B_\cv} = \timederivative{} \iiint_\cv \frac{B}{\vol} \diff \vol  = \timederivative{} \iiint_\cv \rho b \diff \vol \label{eq_secondbit}
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			\end{IEEEeqnarray}
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		Obtaining a value for this integral may be difficult, especially if the volume of the control volume CV is itself a function of time.
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		The term $\dot B_\net$ can be evaluated by quantifying, for each area element $\diff A$ of the control volume’s surface, the surface flow rate $\rho b V_\perp$ of property $B$ that flows through it (\cref{fig_cv_da}). The integral over the entire control volume surface CS of this term is:
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			\begin{IEEEeqnarray}{rCl}
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				\dot B_\net	 = \iint_\cs \rho b V_\perp \diff A = \iint_\cs \rho b \ (\vec V_\rel \cdot \vec n) \diff A \label{eq_thirdbit}
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			\end{IEEEeqnarray}
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			\begin{equationterms}
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				\item where flows and velocities are positive outwards and negative inwards by convention,
				\item 		\tab CV			\tab is the control volume,
				\item 		\tab CS			\tab\tab is the the control surface (enclosing the control volume),
				\item 		\tab $\vec n$ 	\tab\tab\tab is a unit vector on each surface element $\diff A$ pointing outwards,
				\item 		\tab $\vec V_\rel$ 	\tab is the local velocity of fluid relative to the control surface,
				\item and 	\tab $V_\perp \equiv \vec V_\rel \cdot \vec n$ is the local cross-surface speed.
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			\end{equationterms}
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			\begin{figure}
				\begin{center}
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					\includegraphics[width=0.7\textwidth]{concept_vrel_vecn.png}
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				\end{center}
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				\supercaption{Part of the system may be flowing through an arbitrary piece of the control surface with area $\diff A$. The $\vec n$ vector defines the orientation of $\diff A$ surface, and by convention is always pointed outwards.}{\wcfile{System control volume intregral analysis.svg}{Figure} \cczero \oc}
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				\label{fig_cv_da}
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			\end{figure}

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		By inserting equations~\ref{eq_secondbit} and~\ref{eq_thirdbit} into equation~\ref{eq_rtt_basic}, we obtain:
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			\begin{mdframed}\vspace{-0.5cm}\begin{IEEEeqnarray}{rCl}
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				\timederivative{B_\sys} & = & \timederivative{} \iiint_\cv \rho b \diff \vol  +  \iint_\cs \rho b \ (\vec V_\rel \cdot \vec n) \diff A \label{eq_rtt}
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			\end{IEEEeqnarray}
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			\end{mdframed}
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		Equation~\ref{eq_rtt} is named the \vocab{Reynolds’ transport theorem}; it stands now as a general, abstract accounting tool, but as we soon replace $B$ by meaningful variables, it will prove extremely useful, allowing us to quantify the \emph{net} effect of the flow of a system through a volume for which border properties are known.
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		In the following sections we are going to use this equation to assert four key physical principles (\S\ref{ch_conservation_equations}) in order to analyze the flow of fluids:
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		\begin{itemize}
			\item mass conservation;
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			\item change of linear momentum;
			\item change of angular momentum;
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			\item energy conservation.
		\end{itemize}

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\section{Balance of mass}
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	\commonsvideo{Niccolò Paganini - Caprice No.5 - David Hernando.ogv}{https://frama.link/yVesHaSk}{with sufficient skills (and lots of practice!), it is possible for a musician to produce an uninterrupted stream of air into an instrument while still continuing to breathe, a technique called \vocab{circular breathing}. Can you identify the different terms of eq.~\ref{eq_rtt_mass} as they apply to the clarinetist’s mouth?}{David Hernando Vitores (\ccbysa)}
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	In this section, we focus on simply asserting that mass is conserved (eq.~\ref{eq_massconservation} p.\pageref{eq_massconservation}). Our study of the fluid’s properties at the borders of the control volume is made by replacing variable $B$ by mass $m$. Thus $\timederivative{B}$ becomes $\timederivative{m_\sys}$, which by definition is~zero.\\
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	In a similar fashion, $b \equiv B/m = m/m = 1$ and consequently the Reynolds transport theorem (\ref{eq_rtt}) becomes:		
		\begin{IEEEeqnarray}{CCCCCCC}
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			\timederivative{m_\sys} & = & 0 & = & \timederivative{} \iiint_\cv \rho \diff \vol  & + & \iint_\cs \rho \ (\vec V_\rel \cdot \vec n) \diff A \label{eq_rtt_mass}\\\nonumber\\
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			\begin{array}{c} {\scriptstyle \text{the time change}} \\ {\scriptstyle \text{of the system’s mass}} \end{array}	& = &  0	& = & \begin{array}{c} {\scriptstyle \text{the rate of change}} \\ {\scriptstyle \text{of mass inside}} \\ {\scriptstyle \text{the control volume}} \end{array}	& + & \begin{array}{c} {\scriptstyle \text{the net mass flow}} \\ {\scriptstyle \text{at the borders}} \\ {\scriptstyle \text{of the control volume}} \end{array} \nonumber
		\end{IEEEeqnarray}

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	This equation~\ref{eq_rtt_mass} is often called \vocab{continuity equation}. It allows us to compare the incoming and outgoing mass flows through the borders of the control volume.
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	When the control volume has well-defined inlets and outlets through which the term $\rho (\vec V_\rel \cdot \vec n)$ can be considered uniform (as for example in \cref{fig_cv_continuity_simple}), this equation reduces to:
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		\begin{IEEEeqnarray}{rCl}
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			0 	& = & \timederivative{} \iiint_\cv \rho \diff \vol  + \sum_\out \left\{ \rho V_\perp A\right\} + \sum_\inn \left\{ \rho V_\perp A\right\}\label{eq_rtt_mass_simple}\\
				& = & \timederivative{} \iiint_\cv \rho \diff \vol  + \sum_\out \left\{ \rho |V_\perp| A\right\} - \sum_\inn \left\{ \rho |V_\perp| A\right\} \nonumber\\
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				& = & \timederivative{} \iiint_\cv \rho \diff \vol  + \sum_\out \left\{ \dot m \right\} - \sum_\inn \left\{ \dot m \right\}\label{eq_rtt_mass_simple_two}
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		\end{IEEEeqnarray}
		
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	In equation~\ref{eq_rtt_mass_simple}, the term $\rho V_\perp A$ at each inlet or outlet corresponds to the local mass flow~$\pm \dot m$ (positive inwards, negative outwards) through the boundary.
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			\begin{figure}
				\begin{center}
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					\includegraphics[width=0.8\textwidth]{simple_cv_fnet.png}
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				\end{center}
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				\supercaption{A control volume for which the system’s properties are uniform at each inlet and outlet. Here eq.~\ref{eq_rtt_mass} translates as $0 = \timederivative{} \iiint_\cv \rho \diff \vol  + \rho_3 |V_{\perp 3}| A_3 + \rho_2 |V_{\perp 2}| A_2 - \rho_1 |V_{\perp 1}| A_1$.}{\wcfile{Integral analysis angular momentum sketch.svg}{Figure} \cczero \oc}
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				\label{fig_cv_continuity_simple}
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			\end{figure}

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	With equation~\ref{eq_rtt_mass_simple_two} we can see that when the flow is steady (\S\ref{ch_classification_of_fluid_flows}), the last two terms amount to zero, and the integral $\iiint_\cv \rho \diff \vol$ (the total amount of mass in the control volume) does not change with time.\dontbreakpage

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\section{Balance of linear momentum}
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	In this section, we apply Newton’s second law: we assert that the variation of system’s linear momentum is equal to the net force being applied to it (eq.~\ref{eq_secondlaw} p.\pageref{eq_secondlaw}). Our study of the fluid’s properties at the control surface is carried out by replacing variable $B$ by the quantity $m \vec V$: momentum. Thus, $\timederivative{B_\sys}$ becomes $\timederivative{(m \vec V_\sys)}$, which is equal to $\vec F_\net$, the vector sum of forces applied on the system as it transits the control volume.\\
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	In a similar fashion, $b \equiv B/m = \vec V$ and equation~\ref{eq_rtt}, the Reynolds transport theorem, becomes:
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		\begin{IEEEeqnarray}{CCCCCCC}
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			\timederivative{(m \vec V_\text{sys})} & = & \vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol  & + & \iint_\cs \rho \vec V \ (\vec V_\rel \cdot \vec n) \diff A \nonumber\\\label{eq_rtt_linearmom}\\
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			& & \begin{array}{c} {\scriptstyle \text{the vector sum of}} \\ {\scriptstyle \text{forces on the system}} \end{array}	& = & \begin{array}{c} {\scriptstyle \text{the rate of change}} \\ {\scriptstyle \text{of linear momentum}} \\ {\scriptstyle \text{within the control volume}} \end{array}	& + & \begin{array}{c} {\scriptstyle \text{the net flow of linear momen-}} \\ {\scriptstyle \text{tum through the boundaries}} \\ {\scriptstyle \text{of the control volume}} \end{array} \nonumber
		\end{IEEEeqnarray}

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	When the control volume has well-defined inlets and outlets through which the term $\rho \vec V (\vec V_\rel \cdot \vec n)$ can be considered uniform (\cref{fig_cv_continuity_simple}), this equation reduces to:	
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		\begin{IEEEeqnarray}{rCl}
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			\vec F_\net & = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol  + \sum_\out \left\{ (\rho |V_\perp| A) \vec V\right\} - \sum_\inn \left\{ (\rho |V_\perp| A) \vec V\right\} \label{eq_rtt_linearmom_simple}\\
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				& = & \timederivative{} \iiint_\cv \rho \vec V \diff \vol  + \sum_\out \left\{ \dot m \vec V\right\} - \sum_\inn \left\{ \dot m \vec V\right\}\label{eq_rtt_linearmom_simple_two}
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		\end{IEEEeqnarray}
		
			\begin{figure}
				\begin{center}
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					\includegraphics[width=0.8\textwidth]{simple_cv_fnet.png}
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				\end{center}
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				\supercaption{The same control volume as in \cref{fig_cv_continuity_simple}. Here, since the system’s properties are uniform at each inlet and outlet, eq.~\ref{eq_rtt_linearmom} translates as $\vec F_\net = \timederivative{} \iiint_\cv \rho \vec V \diff \vol  + \rho_3 |V_{\perp 3}| A_3 \vec V_3 + \rho_2 |V_{\perp 2}| A_2 \vec V_2 - \rho_1 |V_{\perp 1}| A_1 \vec V_1$.}{\wcfile{Integral analysis angular momentum sketch.svg}{Figure} \cczero \oc}
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				\label{fig_cv_linearmom_simple}
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			\end{figure}

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	Let us observe the four terms of equation~\ref{eq_rtt_linearmom_simple_two} for a moment, for they are full of subtleties.
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	First, we notice that even if the flow is steady (an therefore that $\sum_\net \left(\dot m\right) = \SI{0}{\kilogram\per\second}$ since $\timederivative{} = 0$), the last two terms do not necessarily cancel each other (i.e. it is possible that $\sum_\net \left(\dot m\vec V\right) \neq \vec 0$).
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	The reverse also applies: it is quite possible that $\vec F_\net \neq \vec 0$ even if the net momentum flow through the boundaries is null (that is, even if $\sum_\net \left(\dot m \vec V\right) = \vec 0$). This would be the case, if $\iiint_\cv \rho \diff \vol$ (the total amount of momentum within the borders of the control volume) varies with time. Walking forwards and backwards within a rowboat would cause such an effect.
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	\youtubethumb{S6JKwzK37_8}{as a person walks, the deflection of the air passing around their body can be used to sustain the flight of a paper airplane (a \vocab{walkalong glider}). Can you figure out the momentum flow entering and leaving a control volume surrounding the glider?}{Y:sciencetoymaker (\styl)}
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	From equation~\ref{eq_rtt_linearmom_simple_two} therefore, we read that two distinct phenomena can result in a net force on the system:
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		\begin{itemize}
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			\item A difference between the values of $\dot m \vec V$ of the fluid at the entrance and exit of the control volume (caused, for example, by a deviation of the flow or by a mass flow imbalance);
			\item A change in time of the momentum $m \vec V$ within the control volume (for example, with the acceleration or the variation of the mass of the control volume).
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		\end{itemize}
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	These two factors may cancel each other, so that the system may well be able to travel through the control volume without any net force being applied to~it.
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\clearpage %handmade
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\section{Balance of angular momentum}
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	\youtubethumb{nmEe7Dq01AU}{pre-lecture briefing for this chapter, part 2/2}{\oc (\ccby)}
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	In this third spin on the Reynolds transport theorem, we assert that the change of the angular momentum of a system about a point X is equal to the net moment applied on the system about this point (eq.~\ref{eq_secondlawmom} p.\pageref{eq_secondlawmom}). Our study of the fluid’s properties at the borders of the control volume is made by replacing the variable $B$ by the angular momentum $\vec r_{\X m} \wedge m \vec V$. Thus, $\timederivative{B_\sys}$ becomes $\timederivative{(\vec r_{\X m} \wedge m \vec V_\sys)}$, which is equal to~$\vec M_\net$, the vector sum of moments applied on the system about point X as it transits through the control volume.
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	In a similar fashion, $b \equiv B/m = \vec r \wedge \vec V$ and equation~\ref{eq_rtt}, the Reynolds transport theorem, becomes:
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		\begin{IEEEeqnarray}{cCcCccc}
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			\timederivative{(\vec r_{\X m} \wedge m \vec V)_\sys} & = & \vec M_{\net, \X} & = & \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol  & + & \iint_\cs \vec r_{\X m} \wedge \rho \ (\vec V_\rel \cdot \vec n) \vec V \diff A \nonumber\\\label{eq_rtt_angularmom}\\
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			& & \begin{array}{c} {\scriptstyle \text{the sum of}} \\ {\scriptstyle \text{moments applied}} \\ {\scriptstyle \text{to the system}} \end{array}	& = & \begin{array}{c} {\scriptstyle \text{the rate of change of}} \\ {\scriptstyle \text{the angular momentum}} \\ {\scriptstyle \text{in the control volume}} \end{array}	& + & \begin{array}{c} {\scriptstyle \text{the net flow of angular}} \\ {\scriptstyle \text{momentum through the}} \\ {\scriptstyle \text{control volume’s boundaries}} \end{array} \nonumber
		\end{IEEEeqnarray}
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		\begin{equationterms}
			\item in which $\vec r_{\X m}$ is a vector giving the position of any mass $m$ relative to point $X$.
		\end{equationterms}
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	\youtubethumb{4cvGGxTsQx0}{rocket landing gone wrong. Can you compute the moment exerted by the top thruster around the base of the rocket as it (unsuccessfully) attempts to compensate for the collapsed landing leg?}{Y:SciNews (\styl)} 
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	When the control volume has well-defined inlets and outlets through which the term $\vec r_{\X m} \wedge \rho \ (\vec V_\rel \cdot \vec n) \vec V$ can be considered uniform (\cref{fig_cv_angularmomentum_simple}), this equation reduces~to:
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		\begin{IEEEeqnarray}{rCCCCCl}
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			\vec M_{\net, \X} & = & \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol  &+& \sum_\out \left\{ \vec r_{\X m} \wedge \dot m \vec V\right\} - \sum_\inn \left\{ \vec r_{\X m} \wedge \dot m \vec V\right\} \nonumber\\\label{eq_rtt_angularmom_simple}
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		\end{IEEEeqnarray}
		
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			\begin{figure}[ht!]%handmade
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				\begin{center}
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					\includegraphics[width=0.6\textwidth]{simple_cv_mnet.png}
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				\end{center}
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				\supercaption{A control volume for which the properties of the system are uniform at each inlet or outlet. Here the moment about point X is $\vec M_{\net, \X} \approx \timederivative{} \iiint_\cv \vec r_{\X m} \wedge \rho \vec V \diff \vol  + \vec r_{2} \wedge |\dot m_2| \vec V_2 - \vec r_{1} \wedge |\dot m_1| \vec V_1$.}{\wcfile{Integral analysis angular momentum sketch.svg}{Figure} \cczero \oc}
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				\label{fig_cv_angularmomentum_simple}
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			\end{figure}

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	Equation~\ref{eq_rtt_angularmom_simple} allows us to quantify, with relative ease, the moment exerted on a system based on inlet and outlet velocities of a control volume.
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\dontbreakpage %handmade
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\section{Balance of energy}
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	We conclude our frantic exploration of control volume analysis with the first principle of thermodynamics. We now simply assert that the change in the energy of a system can only be due to well-identified transfers (eq.~\ref{eq_firstprinciple} p.\pageref{eq_firstprinciple}).  Our study of the fluid’s properties at the borders of the control volume is made by replacing variable $B$ by an amount of energy~$E_\sys$. Now, $\timederivative{E_\sys}$ can be attributed to three contributors:
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		\begin{equation}
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			\timederivative{E_\sys} = \dot Q_{\net \ \inn} + \dot W_\text{shaft, net in} + \dot W_\text{pressure, net in}
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		\end{equation}	
	\begin{equationterms}
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		\item where \tab $\dot Q_{\net \ \inn}$ 		\tab is the net power transfered as heat;
		\item 		\tab $\dot W_\text{shaft, net in}$ 	\tab is the net power added as work with a shaft;
		\item and 	\tab $\dot W_\text{pressure, net in}$ 	\tab is the net power required to enter and leave the control volume.
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	\end{equationterms}
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	It follows that $b \equiv B/m = E/m \equiv e$; and $e$ is broken down into
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		\begin{equation}
			e = i + e_k + e_p
		\end{equation}
	\begin{equationterms}
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		\item where \tab $i$ 	\tab\tab is the specific internal energy (\si{\joule\per\kilogram});
		\item 		\tab $e_k$ 	\tab the specific kinetic energy (\si{\joule\per\kilogram});
		\item and 	\tab $e_p$ 	\tab the specific potential energy (\si{\joule\per\kilogram}).
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	\end{equationterms}
	
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	Now, the Reynolds transport theorem (equation~\ref{eq_rtt}) becomes:
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	\begin{adjustwidth}{-3cm}{0cm}%handmade
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		\begin{IEEEeqnarray}{cCcCcCc}
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			\timederivative{E_\sys} & = & \dot Q_{\net \ \inn} + \dot W_\text{shaft, net in} + \dot W_\text{pressure, net in} & = & \timederivative{} \iiint_\cv \rho\ e \diff \vol  & + & \iint_\cs \rho\ e \ (\vec V_\rel \cdot \vec n) \diff A \nonumber\\\label{eq_rtt_energy}
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\section{Limits of integral analysis}
\label{ch_limits_integral_analysis}

		Integral analysis is an incredibly useful tool in fluid dynamics: in any given problem, it allows us to rapidly describe and calculate the main fluid phenomena at hand. The net force exerted on the fluid as it is deflected downwards by a helicopter, for example, can be calculated using just a loosely-drawn control volume and a single vector equation.
		
		As we progress through exercise sheet 3, however, the limits of this method slowly become apparent. They are twofold.
		\begin{itemize}
			\item First, we are confined to calculating the \emph{net} effect of fluid flow. The net force, for example, encompasses the integral effect of all forces —due to pressure, shear, and gravity— applied on the fluid as it transits through the control volume. Integral analysis gives us absolutely no way of distinguishing between those sub-components. In order to do that (for example, to calculate which part of a pump’s mechanical power is lost to internal viscous effects), we would need to look within the control volume.
			\item Second, all four of our equations in this chapter only work in one direction. The value $\diff B_\sys / \diff t$ of any finite integral cannot be used to find which function $\rho b V_\perp \diff A$ was integrated over the control surface to obtain it. For example, there are an \emph{infinite} number of velocity profiles which will result in a net force of \SI{-12}{\newton}. Knowing the net value of an integral, we cannot deduce the conditions which lead to it.\\
			In practice, this is a major limitation on the use of integral analysis, because it confines us to working with large swaths of experimental data gathered at the borders of our control volumes. From the wake below the helicopter, we deduce the net force; but the net force tells us nothing about the shape of the wake.
		\end{itemize}
		Clearly, in order to overcome these limitations, we are going to need to open up the control volume, and look at the details of the flow within — perhaps by dividing it into a myriad of sub-control volumes. This is what we set ourselves to in chapter~4, with a thundering and formidable methodology we shall call \vocab{derivative analysis}. % ta-daaaa!
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\atendofchapternotes